$\lim_{x\to-\infty}\dfrac{\sqrt{16x^4-8x^2}}{x^2-2}=$
Let's find this limit directly. To do that, we will want to divide both the numerator and the denominator by the same quantity, in a way that will help us derive the limit. Since the leading term of the denominator is $x^2$, let's divide by $x^2$. In the numerator, let's divide by $\sqrt{x^4}$, since for any value, $x^2=\sqrt{x^4}$. $\begin{aligned} &\phantom{=}\lim_{x\to-\infty}\dfrac{\sqrt{16x^4-8x^2}}{x^2-2} \\\\ &=\lim_{x\to-\infty}\dfrac{\dfrac{\sqrt{16x^4-8x^2}}{\sqrt{x^4}}}{\dfrac{x^2-2}{x^2}} \gray{\text{Divide sides by }x^2=\sqrt{x^4}} \end{aligned}$ Now let's continue by simplifying the expression and using the fact that for any nonzero number $k$ and positive power $n$, the limit $\lim_{x\to-\infty}\dfrac{k}{x^n}$ is equal to $0$. $\begin{aligned} &=\lim_{x\to-\infty}\dfrac{\sqrt{\dfrac{16\cancel{x^4}}{\cancel{x^4}}-\dfrac{8\cancel {x^2}}{\cancel {x^2}\cdot x^2}}}{\dfrac{1\cancel{x^2}}{\cancel{x^2}}-\dfrac{2}{x^2}} \\\\ &=\lim_{x\to-\infty}\dfrac{\sqrt{16-\dfrac{8}{x^2}}}{1-\dfrac{2}{x^2}} \\\\ &=\lim_{x\to-\infty}\dfrac{\sqrt{16-0}}{1-0} \gray{\lim_{x\to-\infty}\dfrac{k}{x^n}=0} \\\\ &=\dfrac{\sqrt{16}}{1} \\\\ &=\dfrac{4}{1} \\\\ &=4 \end{aligned}$ In conclusion, $\lim_{x\to-\infty}\dfrac{\sqrt{16x^4-8x^2}}{x^2-2}=4$.